In 1987, R. Paris proved that the nested radical expression for $\phi$,

$$\phi=\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\dots}}}}$$

approaches $\phi$ at a constant rate. For example, defining $\phi_n$ as using $n = 5, 6, 7$ "ones" respectively, then,

$$(1/2)(\phi-\phi_5)(2\phi)^5 = 1.0977\dots$$

$$(1/2)(\phi-\phi_6)(2\phi)^6 = 1.0983\dots$$

$$(1/2)(\phi-\phi_7)(2\phi)^7 = 1.0985\dots$$

which is approaching the *Paris constant* $R = 1.09864196\dots$. It seems it can be generalized. Define,

$$x_n=\sqrt[k]{1+\sqrt[k]{1+\sqrt[k]{1+\sqrt[k]{1_n+\dots}}}}\tag{1}$$

for integer $k>1$ and the equations,

$$x^k = x+1\tag{2}$$

$$y = \frac{1}{x}+1\tag{3}$$

where $x$ is the root of $(2)$ such that $x = x_n$ as $n \to \infty$ in $(1)$. Then one can conjecture that,

$$\lim_{n\to\infty}(1/2)(x-x_n)(ky)^n = C_k\tag{4}$$

for some constant $C_k$. The Paris constant is simply the case $C_2$.

I tested it for increasing large $k$. The sequence of $C_k$ seem to be themselves approaching a constant. The rate is very slow, so for *much higher* $k = 10^{14},10^{15},10^{16}$,

$$C_{10^{14}} = 0.6931471805599500\dots$$

$$C_{10^{15}} = 0.6931471805599457\dots$$

$$C_{10^{16}} = 0.6931471805599454\dots$$

Compare to,

$$\ln 2 = 0.6931471805599453\dots$$

*Question:*

- Does $C_k \to \ln 2$, as $k \to \infty$?

$\color{blue}{Edit,\; Nov.\;25}$

More generally, define,

$$x_n=\sqrt[k]{a+\sqrt[k]{a+\sqrt[k]{a+\sqrt[k]{a_n+\dots}}}}\tag{5}$$

for integers $a\ge 1,\;k>1$ and,

$$x^k = x+a\tag{6}$$

$$y = \frac{a}{x}+1\tag{7}$$

Then it seems,

$$\lim_{n\to\infty}(1/2)(x-x_n)(ky)^n = C_{a,k}\tag{8}$$

The Paris constant is the case $C_{1,2}$. Is it true that as $k \to \infty$, then,

$$\lim_{k\to \infty} C_{1,k} = \ln 2$$

$$\lim_{k\to \infty} C_{2,k} = \tfrac{3}{2} \ln \tfrac{3}{2}$$

$$\lim_{k\to \infty} C_{3,k} = \tfrac{4}{2} \ln \tfrac{4}{3}$$

$$\lim_{k\to \infty} C_{4,k} = \tfrac{5}{2} \ln \tfrac{5}{4}$$

and so on?

**P.S.** The only known closed-form in terms of transcendental constants is $C_{2,2} = \pi^2/8$.